Question

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 68 manufacturing companies located in the Southwest. The mean expense is $49.41 million and the standard deviation is $11.45 million. Is it reasonable to conclude the sample data are from a population that follows a normal probability distribution?

Answer 1

Answer:

$\bar X= 49.41$ represent the sample mean

$s= 11.45$ represent the sample deviation

n = 68 represent the sample size

Since the sample size is large enough n>30 we have enough evidence to conclude that the normal approximation for the sample mean makes sense. And the distribution for the sample mean would be given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we have the following data given:

$\bar X= 49.41$ represent the sample mean

$s= 11.45$ represent the sample deviation

n = 68 represent the sample size

Since the sample size is large enough n>30 we have enough evidence to conclude that the normal approximation for the sample mean makes sense. And the distribution for the sample mean would be given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$